Question 332: Electromagnetic Induction

Question

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure.
The potential difference developed across the ring when its speed is $$v$$, is

Accepted Answer

Potential difference that is developed across ring when its speed is v :
$$\varepsilon $$ = B v (l
i
Ã¢ÂÂ l
f
)
where, l
i
Ã¢ÂÂ l
f
= displacement between end of semicircular ring = 2r
Hence, $$\varepsilon $$ = Bv(2r) where R is at high potential.

Answer

zero

Answer

$${{Bv\pi {r^2}} \over 2}$$ and P is at higher potential

Answer

$$\pi rBV$$ and R is at higher potential

Answer

2rBV and R is at higher potential

Electromagnetic Induction

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is $v$, is

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